Q24 of 158 Page 159

In Fig. 6.13, ∠BAC = 90°, AD ⊥ BC and ∠BAD = 50°, then ∠ACD is

Given: ∠BAC = 90°, AD ⊥ BC and ∠BAD = 50°


Formula Used/Theory:-


Sum of all angles of triangle is 180°


As ∠BAC = ∠BAD + ∠CAD


90° = 50° + ∠CAD


∠CAD = 90° - 50° = 40°


In Δ ADC


∠ADC = 90° ∵ AD is perpendicular to BC


∠ADC + ∠CAD + ∠ACD = 180°


90° + 40° + ∠ACD = 180°


130° + ∠ACD = 180°


∠ACD = 180° - 130° = 50°

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